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Find the length of the diagonal, round to the nearest tenth (( please answer, ill give brainliest))

Lilit [14]

Answer:

Length of diagonal in the first top figure is 13 yd

Okay for this one, the formula of a diagonal in a cuboid is the root of sum of squares

length, breadth and height

so

D = root of 12^2 + 4^2 + 3^2

D = root of 144 + 16 + 9

D = root 169

D = 13yd

Length of diagonal in the bottom figure is 10.77 which is 10.8m

For this one, you have to find the diameter of base first

Since radius of the base is 5 diameter = 10

Since its a right angled triangle, Hypotenuse square = sum of sides square

Diagonal^2 = 10^2 + 4^2

D^2 = 100 + 16

D^2 = 116

D = root 116

D = 10.77

D = 10.8 m

6 0

3 years ago

3/4 of Brianna's garden is planted in vegetables. carrots are planted in 1/2 of her vegetable section of the garden. How much of

Artemon [7]

3/8 is the amount of her garden that is planted in carrots.

4 0

3 years ago

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Find g: 3g + 15 − 12 + 3 − 2g + 4 = 1

lisov135 [29]

Answer:Group of answer choices

Step-by-step explanation:

What you first want to do is like and unlike terms so basically liketerms are terms withthe same sign and unlike terms are terms with diferent signs for instinse 3g and 2g are like terms so -33 and -9 and -9 and 9 are unlike terms so you just have to find the like and unlike terms to answer the question!

6 0

3 years ago

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The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas

saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by $s(t) = -t^3+3t^2+7t +4$ , where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

$s(t) = -t^3+3t^2+7t +4$

Derivate w.r.t. time,

$v(t)=s'(t) = -3t^2+6t+7$

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

$x=-\frac{b}{2a}$

Where, a=-3, b=6 and c=7

$t=-\frac{6}{2(-3)}$

$t=-\frac{6}{-6}$

$t=1$

As 1 lie between interval [0,4]

Substitute t=1 in the function,

$v(t)= -3(1)^2+6(1)+7$

$v(t)= -3+6+7$

$v(1)=10$

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0

3 years ago

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1 Solve for the unknown variable:

Burka [1]

Answer:

Approximately 6.4

Step-by-step explanation:

We can use the pythagorean thereom here, that tells us (a^2)+(b^2)=c^2. C is the hypotenuse, the side opposite from the right angle, while a and b are the other sides. We can insert 5 and 4 as a and b, and solve for c

:(5^2)+(4^2)=c^2

:25+16=c^2

:41=c^2

:sqrt(41)=6.4=c (We square rooted both sides. 6.4 is only rounded to the nearest hundredths place.) Hope this helps!

8 0

3 years ago

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